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2 years ago

Calculate the density (in grams per milliliter) of the following gas at 97 °C and 755 mm Hg Sulfur dioxide (SO2)

Chemistry

1 answer:

Diano4ka-milaya [45]2 years ago

3 0

Answer:

Density = 2.09×10⁻³ g/mL

Explanation:

We solve this problem from the Ideal Gases Law.

P . V = n . R . T

We know that in STP conditions 1 mol of any gas occupies 22.4 L. So we can compare both situations.

22.4 L = 22400 mL

(P₁ . V₁) / T₁ = (P₂ . V₂) / T₂

We make some conversions:

97°C + 273 = 370 K

755 mmHg . 1 atm / 760mmHg = 0.993 atm

We replace data: (1 atm . 22400 mL) / 273 K = (0.993 atm . V₂) / 370 K

((1 atm . 22400 mL) / 273 K) . 370K = 0.993 atm . V₂

V₂ = 30359 atm . mL / 0.993 atm → 30573 mL

Under those conditions (97 °C and 755 mm Hg) 1 mol of gas is contained in 30573 mL.

We determine the amount of gas: 1 mol of SO₂ weighs 64.06g

Density = m/V → 64.06 g / 30573 mL = 2.09×10⁻³ g/mL

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Explanation:

For this question we have to keep in mind that the unknown substance is a gas, therefore we can use the ideal gas law:

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So, we can solve for "n" (moles):

$1~atm*3.16~L~=~n*0.082~\frac{atm*L}{mol*K}*305.15~K$

$n=\frac{1~atm*3.16~L~}{0.082~\frac{atm*L}{mol*K}*305.15~K}$

$n=0.126~mol$

Now, we have to remember that the molar mass value has "g/mol" units. We already have the grams (9.33 g), so we have to divide by the moles:

$molar~mass=\frac{9.33~grams}{0.126~mol}$

$molar~mass=73.88\frac{grams}{mol}$

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